03/20/2018, 12:16 AM

In the title I mentioned that this is a repost.

Indeed these ideas are not completely new and have been mentioned before.

However I Will reinvent my own wheel , by a different presentation of the/my ideas ...

The basic principle is very old : infinite Sum and telescoping.

We want to solve t(exp(x)) = e t(x).

Define

0 < a < 1.

f(x,a) = exp(x) + a x.

f^[n+1](x,a) = f( f^[n](x,a) ,a).

—-

Now consider

for Every real y there is a solution x to

f(x,a) = y.

Nomatter What a is , but still 0 < a < 1.

——

Now define the Sum from m = - Oo to Oo :

t(z,a) = ... + exp(z + 2)/f^[-2](1/2,a) + exp(z + 1)/f^[-1](1/2,a) + ...

So t(z,a) = sum_m exp(z - m)/f^[m](1/2,a).

Now let Lim a —> 0 :

z = x then :

Lim t(z,0) = t(z) = t(x)

t(exp(x)) = e t(x).

Qed

___

For practical reasons I suggest

Compute for x < 1 :

t(x) by computing t(exp(x))/e.

——

We can continue

since 1/z + 1/exp(z) + 1/exp(exp(z)) + ...

Is nowhere analytic since exp iterations are chaotic we have :

t(x) is nowhere analytic.

Also

t(exp(x)) = t(x) e.

ln t( exp(x) ) = ln t(x) + 1

Thus

ln t(x) = slog(x).

A nowhere analytic slog(x) !

——-

Due to the telescoping infinite Sum , this mzthod is probably the simplest nowhere analytic solution in the spirit of calculus.

I assume it was worthy of Being repeated.

Regards

Tommy1729

Indeed these ideas are not completely new and have been mentioned before.

However I Will reinvent my own wheel , by a different presentation of the/my ideas ...

The basic principle is very old : infinite Sum and telescoping.

We want to solve t(exp(x)) = e t(x).

Define

0 < a < 1.

f(x,a) = exp(x) + a x.

f^[n+1](x,a) = f( f^[n](x,a) ,a).

—-

Now consider

for Every real y there is a solution x to

f(x,a) = y.

Nomatter What a is , but still 0 < a < 1.

——

Now define the Sum from m = - Oo to Oo :

t(z,a) = ... + exp(z + 2)/f^[-2](1/2,a) + exp(z + 1)/f^[-1](1/2,a) + ...

So t(z,a) = sum_m exp(z - m)/f^[m](1/2,a).

Now let Lim a —> 0 :

z = x then :

Lim t(z,0) = t(z) = t(x)

t(exp(x)) = e t(x).

Qed

___

For practical reasons I suggest

Compute for x < 1 :

t(x) by computing t(exp(x))/e.

——

We can continue

since 1/z + 1/exp(z) + 1/exp(exp(z)) + ...

Is nowhere analytic since exp iterations are chaotic we have :

t(x) is nowhere analytic.

Also

t(exp(x)) = t(x) e.

ln t( exp(x) ) = ln t(x) + 1

Thus

ln t(x) = slog(x).

A nowhere analytic slog(x) !

——-

Due to the telescoping infinite Sum , this mzthod is probably the simplest nowhere analytic solution in the spirit of calculus.

I assume it was worthy of Being repeated.

Regards

Tommy1729